\newproblem{lay:6_5_21}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.5.21}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be an $m\times n$ matrix whose columns are linearly independent. [\textit{Careful}: $A$ need not be square.]
	\begin{enumerate}[a.]
		\item Use Exercise 6.5.19 to show that $A^TA$ is an invertible matrix.
		\item Explain why $A$ must have at least as many rows as columns.
		\item Determine the rank of $A$.
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}
		\item If the columns of $A$ are linearly independent, then the only solution of the problem
		      \begin{center}
						$A\mathbf{x}=\mathbf{0}$
					\end{center}
					is $\mathbf{x}=\mathbf{0}$, that is, $\mathrm{Nul}\{A\}=\{\mathbf{0}\}$ and by Exercise 6.5.19, $\mathrm{Nul}\{A^TA\}=\{\mathbf{0}\}$. By the Invertible
					Matrix Theorem (see Section 2.9) this implies that $A^TA$ is invertible.
		\item $A$ has at least as many rows as columns if $m\geq n$. Note that $A^TA$ is of size $n\times n$ and we need its rank to be $n$ (so that it can be inverted).
		      The rank of a matrix meets:
		      \begin{center}
						$\mathrm{Rank}\{A^TA\}=\mathrm{Rank}\{A\}=\mathrm{Rank}\{A^T\}=\mathrm{Rank}\{AA^T\}$
					\end{center}
					Note also that the rank of $A$ is at most the minimum between $m$ and $n$, so if $A^TA$ is invertible, it must be $m\geq n$ because otherwise the rank of $A$
					would be $m<n$ and $A^TA$ would not be invertible.
		\item See response to previous point, $\mathrm{Rank}\{A\}=n$.
	\end{enumerate}
}
\useproblem{lay:6_5_21}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
